Q22 of 32 Page 20

Let a, b, c, k, be a rational number such that k is not a perfect cube. If

a + b k1/3 + ck2/3 = 0. Then prove that a = b = c = 0.

Given: a + bk1/3 + ck2/3 = 0 ……(i)


a + bk1/3 = -ck2/3


Cubing both sides,


a3 + b3k + 3a2bk1/3 + 3ab2k1/3 = -c3k2


(a3 + b3 + c3k2) + (3a2b)k1/3 + (3ab2)k2/3 = 0


Comparing coefficient to get :


b2 = ac


b2k + c3k2 = 2a2


as b2 < 4ac, discriminant of (i) is negative.


Hence, a = 0, b = 0, c = 0, as no solution.


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